Question

помогите пж кто шарит за алгебру 9 класс

Answer 1

Ответ:

$a) D(f) = [0; \frac{2}{3}] \\\\b) D(f) = [-4; 6) U(6;+ \infty)\\\\c) D(f) = [0; \frac{6}{5}]$

Объяснение:

$a) \sqrt{2x - 3x^2} \\2x - 3x^2 \geq 0\\x(2-3x)\geq 0\\x \geq 0; 2-3x \geq 0\\x \geq 0; 3x \leq 2\\x\geq 0; x \leq \frac{2}{3} \\D(f) = [0; \frac{2}{3}] \\\\b) \sqrt{x^2+6x+8}:{3x+18} \\1) 3x + 18 \neq 0\\x \neq 6\\2) x^2 + 6x + 8 \geq 0\\x\geq -4; x \geq -2\\D(f) = [-4; 6) U(6;+ \infty)\\\\c) y = \sqrt{7x - x^2} + \sqrt{6-5x} \\1) 7x-x^2 \geq 0\\x(7-x)\geq 0\\x \geq 0; x \leq 7\\2) 6-5x \geq 0\\x \leq \frac{6}{5} \\D(f) = [0; \frac{6}{5}]$